Answer
$62.5\;\rm \mu J$
Work Step by Step
First, we need to find the charge stored in the capacitor before connecting it to the two resistors.
$$\Delta V_C=\dfrac{Q}{C}$$
So,
$$Q=C\Delta V_C\tag 1$$
After connecting the capacitor in series to the two resistors, all its stored energy will be dissipated there.
This energy is given by
$$U_C=\dfrac{Q^2}{2C}\tag 2$$
We know that one resistor ($100\;\Omega$) is four times the other ($25\;\Omega$),
$$R_2=4R_1\tag 3$$
and the energy dissipated in the resistor is given by
$$U_{\rm diss}=I^2R\tag 4$$
The current is a function in time but it must be the same in both resistors, so
$$$$
Hence,
$$\dfrac{(U_{\rm diss})_2}{(U_{\rm diss})_1}=\dfrac{I^2R_2}{I^2R_1}=4$$
$$(U_{\rm diss})_2=4(U_{\rm diss})_1\tag 5$$
$$(U_{\rm diss})_1+(U_{\rm diss})_2=U_C$$
Plug from (2),
$$(U_{\rm diss})_1+(U_{\rm diss})_2=\dfrac{Q^2}{2C}$$
Plug from (1),
$$(U_{\rm diss})_1+(U_{\rm diss})_2=\dfrac{C^2(\Delta V_C)^2}{2C}$$
$$(U_{\rm diss})_1+(U_{\rm diss})_2=\frac{1}{2 }C (\Delta V_C)^2$$
Plug from (5),
$$(U_{\rm diss})_1+4(U_{\rm diss})_1=\frac{1}{2 }C (\Delta V_C)^2$$
$$5(U_{\rm diss})_1=\frac{1}{2 }C (\Delta V_C)^2$$
$$ (U_{\rm diss})_1=\frac{1}{10 }C (\Delta V_C)^2$$
Plug the known;
$$ (U_{\rm diss})_1=\frac{1}{10 }(0.25\times 10^{-6}) (50)^2$$
$$ (U_{\rm diss})_1=\color{red}{\bf 6.25\times 10^{-5}}\;\rm J$$
