Answer
$\rm 6\;V,\;8\;V$
Work Step by Step
The peak voltage for the resistor is given by
$$V_R=IR\tag 1$$
and for the capacitor is given by
$$V_C=IX_C$$
where $X_C=1/\omega C=1/2\pi f C$, so
$$V_C= \dfrac{I}{2\pi f C}\tag 2$$
Now we need to find the peal current $I$ which is given by
$$I=\dfrac{\varepsilon_0}{Z}$$
where $\varepsilon_0$ is the peak resource voltage, and $Z=\sqrt{X_C^2+R^2}$, so
$$I=\dfrac{\varepsilon_0}{\sqrt{X_C^2+R^2}}$$
$$I=\dfrac{\varepsilon_0}{\sqrt{\left[\dfrac{1}{2\pi f C}\right]^2+R^2}}\tag 3$$
Plug (3) into (1) to get $V_R$;
$$V_R=\dfrac{\varepsilon_0R}{\sqrt{X_C^2+R^2}} $$
Plug the known;
$$V_R=\dfrac{(10)(150)} {\sqrt{\left[\dfrac{1}{2\pi (10\times 10^3)(80\times 10^{-9})}\right]^2+(150)^2}} $$
$$V_R=\color{red}{\bf 6.02}\;\rm V$$
Plug (3) into (2) to get $V_C$;
$$V_C= \dfrac{1}{2\pi f C} \dfrac{\varepsilon_0}{\sqrt{\left[\dfrac{1}{2\pi f C}\right]^2+R^2}}$$
Plug the known;
$$V_C= \dfrac{1}{2\pi (10\times 10^3)(80\times 10^{-9})} \dfrac{10}{\sqrt{\left[\dfrac{1}{2\pi (10\times 10^3)(80\times 10^{-9})}\right]^2+(150)^2}}$$
$$V_C=\color{red}{\bf 7.98}\;\rm V$$
