Answer
${\bf 3.15}\rm\;\mu C$, ${\bf 5.53}\rm\;\mu C$
Work Step by Step
When the two capacitors are in parallel, the equivalent capacitance is given by
$$C_{eq,1}=C_1+C_2\tag 1$$
And when the two capacitors are in series, the equivalent capacitance is given by
$$C_{eq,2}=\left[\dfrac{1}{C_1}+\dfrac{1}{C_2}\right]^{-1}=\dfrac{C_1C_2}{C_1+C_2}\tag 2$$
So the rms current, in the first case, is given by
$$I_{\rm rms,1}=\dfrac{\varepsilon_{\rm rms}}{X_{C}}=\dfrac{\varepsilon_{\rm rms}}{1/\omega C_{eq,1}}$$
Plug from (1)
$$I_{\rm rms,1}=2\pi f \varepsilon_{\rm rms} (C_1+C_2)$$
So,
$$C_1+C_2=\dfrac{I_{\rm rms,1}}{2\pi f \varepsilon_{\rm rms}}$$
Plug the known;
$$C_1+C_2=\dfrac{(545\times 10^{-3})}{2\pi (1\times 10^3)(10) }$$
$$C_1+C_2=\dfrac{109}{4\times 10^6\pi}\tag 3$$
And the rms current, in the second case, is given by
$$I_{\rm rms,2}=\dfrac{\varepsilon_{\rm rms}}{X_{C}}=\dfrac{\varepsilon_{\rm rms}}{1/\omega C_{eq,2}}$$
Plug from (2);
$$I_{\rm rms,2}=2\pi f \varepsilon_{\rm rms} \dfrac{C_1C_2}{C_1+C_2}$$
So,
$$\dfrac{C_1C_2}{C_1+C_2} =\dfrac{I_{\rm rms,2}}{2\pi f \varepsilon_{\rm rms} }$$
$$ C_1C_2 =(C_1+C_2)\dfrac{I_{\rm rms,2}}{2\pi f \varepsilon_{\rm rms}}$$
Plug the known and plug from (3);
$$ C_1C_2 =\dfrac{109}{4\times 10^6\pi}\cdot \dfrac{ (126\times 10^{-3})}{2\pi (1\times 10^3) (10) }$$
$$ C_1C_2 =\dfrac{6867}{4\times 10^{13}\pi^2} $$
Hence,
$$ C_2 =\dfrac{6867}{4\times 10^{13}\pi^2C_1} $$
Plug into (3);
$$C_1+\dfrac{6867}{4\times 10^{13}\pi^2C_1} =\dfrac{109}{4\times 10^6\pi} $$
Thus, the two capacitors are:
$$C_1=3.15\times 10^{-6}\;\rm C=\color{red}{\bf 3.15}\;\mu C$$
Or,
$$C_2=5.53\times 10^{-6}\;\rm C=\color{red}{\bf 5.53}\;\mu C$$