The equivalent mass of H3BO3 (M = Molar mass of H3BO3) in its reaction with NaOH to from Na2B4O7 is equal to –

(1) M

(2) $\frac{\text{M}}{\text{2}}$

(3) $\frac{\text{M}}{4}$

(4) $\frac{\text{M}}{6}$