Correct Answer is Option (B)

When only galvanometer G is present with the resistance R,



Here I_G = $\frac{V_E}{R+G}$

When shunt of resistance S is connected parallel to galvanometer,



Here I = $\frac{V_E}{R+\frac{GS}{G+S}}$

As deflection is half, here current through galvanometer,

I_G' = $\frac{{\mathrm{I}}_G}{2}$

As both Galvanometer and shunt are parallel then potential are parallel then potential difference same.

$\therefore$   I_G' (G) = (I $-$ I_G^')S

$\Rightarrow$   I'_G (G + S) = IS

$\Rightarrow$   $\frac{{\mathrm{I}}_G}{2}$ = $\frac{\mathrm{I}S}{G+S}$

$\Rightarrow$   $\frac{V_E}{2\left(R+G\right)}$ = $\frac{V_E}{R+\frac{GS}{G+S}}$ $\times$ $\frac{S}{\left(G+S\right)}$

$\Rightarrow$   $\frac{1}{2\left(R+G\right)}$ = $\frac{G+S}{R(G+S)+GS}$ $\times$ $\frac{S}{\left(G+S\right)}$

$\Rightarrow$   RG + RS + GS = 2RS + 2GS

$\Rightarrow$   RG = RS + GS

$\Rightarrow$    S(R + G) = RG