(a) ByOhm's law,
$$
\begin{aligned}
&\mathrm{V}=\mathrm{IR}, \\
&\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}} \\
&\mathrm{I}=6 \mathrm{~V} / 6 \Omega=1 \mathrm{~A}
\end{aligned}
$$
The current in a conductor and drift velocity of electrons are related as $i=n e \mathrm{Av}_{\mathrm{d}}$ where $\mathrm{V}_{\mathrm{d}}$ is drift speed of electrons and $\mathrm{n}$ is number density of electrons.
$$
I=n e A v_d
$$
So, drift velocity $v_d=\frac{i}{\text { neA }}$
As given that,
$\mathrm{n}$ (number of electron/volume) $=10^{29} / \mathrm{m}^3$
length of circuit $=10 \mathrm{~cm}$, cross-section area $(\mathrm{A})=(1 \mathrm{~mm})^2$, $\mathrm{m}_{\mathrm{e}}=9.1 \times 10^{-31}$
On substituting the values
$$
\begin{aligned}
\mathrm{v}_{\mathrm{d}} &=\frac{1}{10^{29} \times 1.6 \times 10^{-19} \times 10^{-6}} \\
&=\frac{1}{1.6} \times 10^{-4} \mathrm{~m} / \mathrm{s}
\end{aligned}
$$
So, the energy absorbed in the form of $\mathrm{KE}$ is
$$
\begin{gathered}
\mathrm{KE}=\frac{1}{2} \mathrm{~m}_{\mathrm{e}} \mathrm{v}_{\mathrm{d}}^2 \times \mathrm{nAI} \\
=\frac{1}{2} \times 9.1 \times 10^{-31} \times \frac{1}{2.56} \times 10^{29} \times 10^{-8} \times 10^{-6} \times 10^{-1} \\
\text { K.E. }=1.78 \times 10^{-17} \mathrm{~J}
\end{gathered}
$$
(b) Power loss
$$
\begin{aligned}
\mathrm{P}^2 &=\mathrm{I}^2 \mathrm{R} \\
&=6 \times 1^2 \\
&=6 \mathrm{~W}=6 \mathrm{~J} / \mathrm{s}
\end{aligned}
$$
We know that,
$$
\text { (Power loss) } P=\frac{E(\text { K.E. })}{t(\text { time })}
$$
So,
$$
\begin{aligned}
t &=\frac{E}{P} \\
&=\frac{1.78 \times 10^{-17}}{6} \\
&=0.29 \times 10^{-17} \\
& \cong 0.3 \times 10^{-17} \\
&=3 \times 10^{-18} \text { second }
\end{aligned}
$$