A parallel plate capacitor has plate area 100 m$$^2$$ and plate separation of 10 m. The space between the plates is filled up to a thickness 5 m with a material of dielectric constant of 10. The resultant capacitance of the system is $$x$$ pF. The value of $$\varepsilon_0 = 8.85 \times 10^{-12}$$ F m$$^{-1}$$. The value of $$x$$ to the nearest integer is ________.

We have a parallel plate capacitor with plate area $$A = 100 \text{ m}^2$$ and plate separation $$d = 10 \text{ m}$$. A dielectric of thickness $$t = 5 \text{ m}$$ with dielectric constant $$K = 10$$ fills part of the gap. The remaining gap of thickness $$(d - t) = 5 \text{ m}$$ is air.

This system behaves as two capacitors in series: one with dielectric (thickness $$t$$, dielectric constant $$K$$) and one with air (thickness $$d - t$$). The capacitance of the dielectric portion is $$C_1 = \dfrac{K\varepsilon_0 A}{t} = \dfrac{10 \times 8.85 \times 10^{-12} \times 100}{5}$$, and the air portion is $$C_2 = \dfrac{\varepsilon_0 A}{d - t} = \dfrac{8.85 \times 10^{-12} \times 100}{5}$$.

For series combination: $$\dfrac{1}{C} = \dfrac{1}{C_1} + \dfrac{1}{C_2} = \dfrac{t}{K\varepsilon_0 A} + \dfrac{d - t}{\varepsilon_0 A} = \dfrac{1}{\varepsilon_0 A}\left(\dfrac{t}{K} + (d - t)\right)$$.

Substituting: $$\dfrac{t}{K} + (d - t) = \dfrac{5}{10} + 5 = 0.5 + 5 = 5.5 \text{ m}$$.

So $$C = \dfrac{\varepsilon_0 A}{5.5} = \dfrac{8.85 \times 10^{-12} \times 100}{5.5} = \dfrac{885 \times 10^{-12}}{5.5} = 160.9 \times 10^{-12} \text{ F} \approx 161 \text{ pF}$$.

The answer is $$\boxed{161}$$.