As shown in the figure, the resistance of a galvanometer G can be...

Let the resistance of the galvanometer be $G$ and the series resistance be $R_1$.
When the key $K$ is open, the current through the galvanometer is:
$I_1 = \dfrac{V}{R_1 + G}$

When the key $K$ is closed, the total resistance of the circuit becomes:
$R_{eq} = R_1 + \dfrac{G R_2}{G + R_2}$

The total current from the battery is $I = \dfrac{V}{R_{eq}}$. The current through the galvanometer is:
$I_2 = I \times \dfrac{R_2}{G + R_2} = \dfrac{V}{R_1 + \dfrac{G R_2}{G + R_2}} \times \dfrac{R_2}{G + R_2} = \dfrac{V R_2}{R_1(G + R_2) + G R_2}$

In the half-deflection condition, $I_2 = \dfrac{I_1}{2}$:
$\dfrac{V R_2}{R_1(G + R_2) + G R_2} = \dfrac{V}{2(R_1 + G)}$

Cross-multiplying and simplifying:
$2 R_2 (R_1 + G) = R_1(G + R_2) + G R_2$
$2 R_1 R_2 + 2 G R_2 = R_1 G + R_1 R_2 + G R_2$
$R_1 R_2 + G R_2 = R_1 G$
$R_1(G - R_2) = G R_2$
$R_1 = \dfrac{G R_2}{G - R_2}$

Given $G = 6\ \Omega$ and $R_2 = 4\ \Omega$, we can find $R_1$:
$R_1 = \dfrac{6 \times 4}{6 - 4} = \dfrac{24}{2} = 12\ \Omega$

In the half-deflection condition (key $K$ closed), the total equivalent resistance of the circuit is:
$R_{eq} = 12 + \dfrac{6 \times 4}{6 + 4} = 12 + 2.4 = 14.4\ \Omega$

The current through the resistor $R_1$ is the total current from the battery:
$I = \dfrac{V}{R_{eq}} = \dfrac{10}{14.4} = \dfrac{100}{144} = \dfrac{25}{36}\ \text{A}$

Converting the current to mA:
$I = \dfrac{25}{36} \times 1000\ \text{mA} = \dfrac{25000}{36}\ \text{mA} \approx 694.44\ \text{mA}$