full-scale deflection voltage for the galvanometer itself is given by:

$$ V_g = I_g \times R_g = 2 \times 10^{-6} \, \text{A} \times 10 \, \Omega = 2 \times 10^{-5} \, \text{V} $$

To achieve a full-scale deflection of 10 mV, the total resistance $R_v$ of the voltmeter should satisfy:

$$ 10 \times 10^{-3} \, \text{V} = I_g \times (R_g + R_s) $$

Where $R_s$ is the series resistance. Solving for $R_s$, we have:

$$ R_s = \frac{10 \times 10^{-3}}{2 \times 10^{-6}} - 10 = 5000 - 10 = 4990 \, \Omega $$

Thus, the resistance of the voltmeter is:

$$ R_v = R_g + R_s = 10 + 4990 = 5000 \, \Omega $$

To convert the galvanometer into an ammeter of range 1 mA, we need to find the shunt resistance $R_a$. For the galvanometer to show full-scale deflection at 1 mA, the remaining current (beyond the 2 µA for full deflection) should pass through the shunt:

$$ I_{shunt} = 1 \, \text{mA} - 2 \, \mu \text{A} = 0.998 \, \text{mA} $$

The voltage across the galvanometer (and hence the shunt) is:

$$ V = I_g \times R_g = 2 \times 10^{-6} \, \text{A} \times 10 \, \Omega = 2 \times 10^{-5} \, \text{V} $$

Then $R_a$, the shunt resistance, can be calculated using:

$$ R_a = \frac{V}{I_{shunt}} = \frac{2 \times 10^{-5}}{0.998 \times 10^{-3}} \approx 0.02004 \, \Omega $$

Finally, to find the ratio of the resistance of the converted voltmeter to that of the ammeter:

$$ \text{Ratio} = \frac{R_v}{R_a} = \frac{5000}{0.02004} \approx 249800 \approx 250000 $$

Therefore, the ratio is closest to:

Option C: 250000 : 1

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