$\Lambda_m^{\large\circ}$ for $Ba(OH)_2=\Lambda_{Be^{+2}}^{\large\circ}+2\Lambda_{OH^{-}}^{\large\circ}$-----(1)
$\Lambda_m^{\large\circ}$ for $BaCl_2=\Lambda_{NH_4^{+}}^{\large\circ}+2\Lambda_{Cl^{-}}^{\large\circ}$-----(2)
$\Lambda_m^{\large\circ}$ for $(NH_4Cl)=\Lambda_{NH_4^{+}}^{\large\circ}+\Lambda_{Cl^{-}}^{\large\circ}$-----(3)
$\Lambda_m^{\large\circ}$ = $(NH_4OH)=\Lambda ^{\large\circ}_{NH_4^+}+\Lambda ^{\large\circ}_{OH^-}$
$\Rightarrow \large\frac{1}{2}$$equ(1)+equ(3)-\large\frac{1}{2}$$equ(2)$
$\Rightarrow \large\frac{1}{2}$$\times 457.6+129.8-\large\frac{1}{2}$$240.6$
$\Rightarrow 238.3\Omega^{-1}cm^2mol^{-1}$
Hence (a) is the correct answer.