Answer
\begin{array}{|c|c|c|}
\hline
R&I({\;\rm A})& \Delta V ({\;\rm V})\\
\hline
4\;\Omega & \frac{9}{16}=0.56& 2.25 \\
\hline
12\;\Omega & \frac{3}{16}=0.19& 2.25 \\
\hline
5\;\Omega & \frac{3}{4}=0.75& 3.75\\
\hline
24\;\Omega & \frac{1}{4}=0.25& 6\\
\hline
3\;\Omega & 1& 3\\
\hline
\end{array}
Work Step by Step
Let's assume that the current comes out from the 12-V battery and charging the 3-V battery, as you see in the last circuit below.
Applying Kirchhoff's loop law on the last circuit below (d), starting from the left lower corner and moving clockwise.
$$\Delta V_{\rm loop}=12-6I-3-3I=0$$
Hence,
$$I=\bf1.0\;\rm A$$
This is the current that passes through $3\;\Omega$.
From circuit (c),
$$I=I_1+I_2=1\;\rm A\tag 1$$
where
$$8I_1=24I_2$$
Hence,
$$ I_1=3I_2$$
Plug into (1),
$$1=3I_2+I_2 $$
Hence,
$$I_2=\bf 0.25\;\rm A$$and
$$I_1=\bf 0.75\;\rm A$$
From circuit (a), we can see that $I_1$ passes through $5\;\Omega$.
Now we need to find $I_3$ and $I_4$, where
$$I_1=I_3+I_4=\bf 0.75\;\rm A\tag 2$$
where
$$12I_3=4I_4$$
Thus,
$$I_4=3I_3$$
Plug into (2),
$$ I_1=I_3+3I_3= 0.75\;\rm A$$
$$I_3=\bf \frac{3}{16}\;\rm A$$
and
$$I_4=\bf \frac{9}{16}\;\rm A$$
\begin{array}{|c|c|c|}
\hline
R&I({\;\rm A})& \Delta V ({\;\rm V})\\
\hline
4\;\Omega & \frac{9}{16}=0.56& 2.25 \\
\hline
12\;\Omega & \frac{3}{16}=0.19& 2.25 \\
\hline
5\;\Omega & \frac{3}{4}=0.75& 3.75\\
\hline
24\;\Omega & \frac{1}{4}=0.25& 6\\
\hline
3\;\Omega & 1& 3\\
\hline
\end{array}
