Area of the small flat search coil, $A=2c{m}^2=2\times {10}^{-4}{m}^2$
Number of turns on the coil, N = 25
Total charge flowing in the coil, $\mathrm{Q} = 7.5\mathrm{mC}=7.5\times {10}^{-3}\mathrm{C}$
Total resistance of the coil and galvanometer, R = 0.50 $\mathrm{\Omega }$
Induced current in the coil. $\mathrm{I}=\frac{\text{Induced }\mathrm{em}\mathrm{f}\left(\mathrm{e}\right)}{\mathrm{R}} ...\left(1\right)$
Induced emf is given as: $\mathrm{e}=-\mathrm{N}\frac{\mathrm{d}\mathrm{\varnothing }}{\mathrm{dt}} ...\left(2\right)$
Where, d$\mathrm{\Phi }$ = Charge in flux
Combining equations (1) and [2), we get

$\begin{array}{l}\mathrm{I}=-\frac{\mathrm{N}\frac{\mathrm{d\varphi }}{\mathrm{dt}}}{\mathrm{R}}\\ \mathrm{Idt}=-\frac{\mathrm{N}}{\mathrm{R}}\mathrm{d\varphi }\end{array}$                                ...(3)

Initial flux through the coil, ${\mathrm{\Phi }}_{\mathrm{i}}$ = BA
Where B = Magnetic field strength
Final flux through the coil, ${\mathrm{\Phi }}_{\mathrm{f}}$ = 0
Integrating equation (3) on both sides, we have

$\int \mathrm{Idt}=\frac{-\mathrm{N}}{\mathrm{R}}{\int }_{\mathrm{d}}^{{\mathrm{\varphi }}_1}\mathrm{d\varphi }$
$\begin{array}{l}\text{ But total charge, }\mathrm{Q}=\int \mathrm{Idt}\\ \therefore \mathrm{Q}=\frac{-\mathrm{N}}{\mathrm{R}}({\mathrm{\varphi }}_{\mathrm{j}}-{\mathrm{\varphi }}_{\mathrm{i}})=\frac{-\mathrm{N}}{\mathrm{R}}(-{\mathrm{\varphi }}_{\mathrm{i}})=+\frac{{\mathrm{N\varphi }}_{\mathrm{i}}}{\mathrm{R}}\\ \mathrm{Q}=\frac{\mathrm{NBA}}{\mathrm{R}}\\ \therefore \mathrm{B}=\frac{\mathrm{QR}}{\mathrm{NA}}\\ =\frac{7.5\times {10}^{-3}\times 0.5}{25\times 2\times {10}^{-4}}=0.75\mathrm{T}\end{array}$

Hence, the field strength of the magnet is 0.75 T.