Hint: The deviation of light depends on the refractive index of the object.

Step 1: Use Snell's law.

Let us consider two planes at r and r+dr. Let the light be incident at an angle $\mathrm{\theta }$ at the plane at r and leave r + dr at an angle $\mathrm{\theta }$+d$\mathrm{\theta }$. Then from Snell's law,

n(r)sin$\mathrm{\theta }$= n(r + dr) sin($\mathrm{\theta }$ + d$\mathrm{\theta }$)
$\Rightarrow$ n(r) sin$\mathrm{\theta }$ = $\left(\mathrm{n}\left(\mathrm{r}\right)+\frac{\mathrm{dn}}{\mathrm{dr}}\mathrm{dr}\right)(\mathrm{sin\theta cosd\theta }+\mathrm{cos\theta } \mathrm{sind\theta })$ 

As $\mathrm{d\theta } \mathrm{is} \mathrm{small}, \mathrm{cosd\theta }=1 \mathrm{and} \mathrm{sind\theta }=\mathrm{d\theta },$

$\Rightarrow$n(r) sin$\mathrm{\theta }$= $\left(\mathrm{n} \left(\mathrm{r}\right)+\frac{\mathrm{dn}}{\mathrm{dr}}\mathrm{dr}\right)(\mathrm{sin\theta }+\mathrm{cos\theta d\theta })$

Ignoring the product of differentials;
                                 $\mathrm{n}\left(\mathrm{r}\right) \mathrm{sin\theta }=\mathrm{n}\left(\mathrm{r}\right)\mathrm{sin\theta }+\frac{\mathrm{dn}}{\mathrm{dr}}\mathrm{dr} \mathrm{sin\theta }+\mathrm{n}\left(\mathrm{r}\right) \mathrm{cos\theta d\theta }$
Or we have,                 $-\frac{\mathrm{dn}}{\mathrm{dr}}\mathrm{tan\theta }=\mathrm{n}\left(\mathrm{r}\right)\frac{\mathrm{d\theta }}{\mathrm{dr}}$
                                 $\frac{2\mathrm{GM}}{{\mathrm{r}}^2{\mathrm{c}}^2}\mathrm{tan\theta }=\left(1+\frac{2\mathrm{GM}}{{\mathrm{rc}}^2}\right)\frac{\mathrm{d\theta }}{\mathrm{dr}}=\frac{\mathrm{d\theta }}{\mathrm{dr}} \left(\mathrm{As} \frac{2\mathrm{GM}}{{\mathrm{rc}}^2}<<<1\right)$
                                ${\int }_0^{{\mathrm{\theta }}_0}\mathrm{d\theta }=\frac{2\mathrm{GM}}{{\mathrm{rc}}^2}{\int }_{-\infty }^{\infty }\frac{\mathrm{tan\theta dr}}{{\mathrm{r}}^2}$

Step 2: Integrate both sides.
Now, substitution for integrals, we have,
Now,                                     ${\mathrm{r}}^2={\mathrm{x}}^2+{\mathrm{R}}^2 \mathrm{and} \tan \mathrm{\theta }=\frac{\mathrm{R}}{\mathrm{x}}$
                                                2rdr= 2xdx
                                        ${\int }_0^{{\mathrm{\theta }}_0}\mathrm{d\theta }=\frac{2\mathrm{GM}}{{\mathrm{c}}^2}{\int }_{-\infty }^{\infty }\frac{\mathrm{xdx}}{{({\mathrm{x}}^2+{\mathrm{R}}^2)}^{{\displaystyle \frac{3}{2}}}}$
Put                                               $\mathrm{x}= \mathrm{R} \mathrm{tan\varphi }$
                                                  $\mathrm{dx}=\mathrm{R} {\sec }^2\mathrm{\varphi d\varphi }$
$\therefore$                                              ${\mathrm{\theta }}_0=\frac{2\mathrm{GMR}}{{\mathrm{c}}^2}{\int }_{-\mathrm{\pi }/2}^{\mathrm{\pi }/2}\frac{{\mathrm{Rsec}}^2\mathrm{\varphi d\varphi }}{{\mathrm{R}}^3 {\sec }^3\mathrm{\varphi }}$
                                                     $=\frac{2\mathrm{GM}}{{\mathrm{Rc}}^2}{\int }_{-\mathrm{\pi }/2}^{\mathrm{\pi }/2}\cos \mathrm{\varphi d} \mathrm{\varphi }=\frac{4 \mathrm{GM}}{{\mathrm{Rc}}^2}$
This is the required proof.