But the answer is 200j

Voltage $V = 100\,\mathrm{V}$, capacitive impedance $Z = -j\,0.5\,\Omega$.
Substitute into the formula:

$$I = \frac{100\,\mathrm{V}}{-j\,0.5\,\Omega}$$

Multiply numerator and denominator by the conjugate of the denominator or rewrite:

$$\frac{1}{-j} = \frac{1}{-j} \cdot \frac{j}{j} = \frac{j}{(-j)j} = \frac{j}{-j^2} = \frac{j}{1} = j$$

So,

$$I = \frac{100\,\mathrm{V}}{0.5\,\Omega} \cdot j = 200\,\mathrm{A} \cdot j$$

Final answer: $\boxed{200j\,\mathrm{A}}$