Answer
See the detailed answer below.
Work Step by Step
For $RC$ filter circuit, we know that
$$V_R=IR=\dfrac{\varepsilon_0R}{\sqrt{R^2+X_C^2}} $$
where $X_C=1/\omega C$;
$$V_R= \dfrac{\varepsilon_0R}{\sqrt{R^2+\dfrac{1}{\omega^2C^2}}}\tag 1$$
and that
$$V_C=IX_C=\dfrac{\varepsilon_0X_C}{\sqrt{R^2+X_C^2}}$$
So,
$$V_C =\dfrac{1}{\omega C}\dfrac{\varepsilon_0 }{\sqrt{R^2+\dfrac{1}{\omega^2C^2}}}\tag 2$$
At $\omega=\omega_c$, where $\omega_c=\dfrac{1}{RC}=\omega\Rightarrow$
$$ \dfrac{1}{\omega C}=R\tag 3$$
Plug (3) into (1),
$$V_R= \dfrac{\varepsilon_0R}{\sqrt{R^2+R^2}} =\dfrac{\varepsilon_0 \color{red}{\bf\not} R}{\sqrt{2}\; \color{red}{\bf\not} R}$$
$$\boxed{V_R =\dfrac{\varepsilon_0 }{\sqrt{2} }}$$
Plug (3) into (2),
$$V_C =R\dfrac{\varepsilon_0 }{\sqrt{R^2+R^2}} $$
$$\boxed{V_C = \dfrac{\varepsilon_0 }{\sqrt{2}} }$$
Thus, from the two boxed formulas, $V_C=V_R$.
$$\boxed{V_C =V_R= \dfrac{\varepsilon_0 }{\sqrt{2}} }$$
