1. Initial voltage across the capacitor

The capacitor has $C = 3\,\mu F$ and initial charge $Q = 30\,\mu C$ so its initial potential difference is $V_0 = \frac{Q}{C} = \frac{30\,\mu C}{3\,\mu F} = 10\,V.$

2. Current at the instant the key is closed

At $t=0^+$, the capacitor behaves like a source of emf equal to its initial voltage $10\,V$.

From the figure, the $5\,M\Omega$ resistor is directly across the capacitor at the moment of closing, so the current through it is $I = \frac{V_0}{R} = \frac{10}{5\times 10^6} \text{ A}.$

Thus, $I = 2\times 10^{-6}\,A = 2\,\mu A.$

So, $x = 2.$

3. Comparison with stored answer

Derived answer: $2$

Stored correct answer: $2$

Hence, the derived answer agrees with the stored answer.