a
For a balance Wheatstone bridge.
$\frac{A}{B} = \frac{D}{C} \Rightarrow \frac{{10}}{5} \ne \frac{4}{4}$ (Unbalanced)
$\frac{{A'}}{B} = \frac{D}{C} \Rightarrow \frac{{A'}}{5} = \frac{4}{4}$$ \Rightarrow $ $A' = 5\,\Omega $
$A'$ $(5\,\Omega )$ is obtained by connecting a $10\,\Omega $ resistance in parallel with $A$.