In the given electrical network, the value of resistance ' $R$ ' when the current in the galvanometer will be zero, is

With no current through the galvanometer, the node labeled $A$ has potential $V_A = 6 \text{ V}$ since the galvanometer connects to the $6 \text{ V}$ battery's positive terminal.

Currents leaving node $A$ are through $R$ and the $4\Omega$ resistor:
$I_R = \frac{V_A}{R} = \frac{6}{R}$
$I_{4\Omega} = \frac{V_A - 10}{4} = \frac{6 - 10}{4} = -1\text{ A}$

Applying KCL at node $A$:
$\frac{6}{R} + (-1) = 0$
$\frac{6}{R} = 1$
$R = 6\Omega$

The resistance is $6\Omega$, corresponding to option B.