1. Check Statement I

A uniform wire has resistance $80\,\Omega$ and is cut into 4 equal parts.

For a uniform wire, resistance is proportional to length: $R \propto l$ So resistance of each part is $R_{\text{each}} = \frac{80}{4} = 20\,\Omega$

Now 4 resistors of $20\,\Omega$ each are connected in parallel. For $n$ equal resistors $R$ in parallel, $R_{\text{eq}} = \frac{R}{n}$ Thus, $R_{\text{eq}} = \frac{20}{4} = 5\,\Omega$

So, Statement I is correct.

2. Check Statement II

Two resistances $2R$ and $3R$ are connected in parallel.

In parallel, potential difference across each resistor is same, say $V$. Thermal energy developed per unit time is power: $P = \frac{V^2}{R}$

So for resistor $3R$, $P_{3R} = \frac{V^2}{3R}$

For resistor $2R$, $P_{2R} = \frac{V^2}{2R}$

Therefore, $P_{3R} : P_{2R} = \frac{1}{3R} : \frac{1}{2R} = 2:3$

Hence thermal energy developed in $3R$ and $2R$ is in the ratio $2:3$ not $3:2$.

So, Statement II is incorrect.

3. Final conclusion

• Statement I: Correct
• Statement II: Incorrect

Therefore, the correct option is: $\boxed{\text{C}}$

4. Comparison with stored answer

Stored correct answer: $\text{C}$

Our derived answer matches the stored answer.