1. Given data

• Number of moles: $n = 5$
• Temperature: $T = 293\,\text{K}$
• Initial pressure: $P_1 = 2.1\,\text{MPa}$
• Final pressure: $P_2 = 1.3\,\text{MPa}$
• Gas is ideal and the process is isothermal
• External pressure given as $4.3\,\text{MPa}$
• $R = 8.314\,\text{J mol}^{-1}\text{K}^{-1}$

We need heat transferred per mole in kJ mol$^{-1}$.

2. Use first law of thermodynamics

For an ideal gas in an isothermal process,

$\Delta U = 0$

So,

$q = -w$

Hence we first calculate the work.

3. Work done against constant external pressure

For an irreversible expansion/compression against constant external pressure,

$w = -P_{\text{ext}}(V_2 - V_1)$

Now for an ideal gas at constant temperature,

$V = \frac{nRT}{P}$

So,

$V_1 = \frac{nRT}{P_1}, \qquad V_2 = \frac{nRT}{P_2}$

Thus,

$V_2 - V_1 = nRT\left(\frac{1}{P_2} - \frac{1}{P_1}\right)$

Therefore,

$w = -P_{\text{ext}}\,nRT\left(\frac{1}{P_2} - \frac{1}{P_1}\right)$

4. Substitute values

First calculate $nRT$:

$nRT = 5 \times 8.314 \times 293 = 12179.01\,\text{J}$

Now,

$\frac{1}{P_2} - \frac{1}{P_1} = \frac{1}{1.3} - \frac{1}{2.1}$

Using MPa consistently,

$\frac{1}{1.3} \approx 0.76923, \qquad \frac{1}{2.1} \approx 0.47619$

So,

$\frac{1}{P_2} - \frac{1}{P_1} \approx 0.29304\,\text{MPa}^{-1}$

Now multiply by $P_{\text{ext}} = 4.3\,\text{MPa}$:

$P_{\text{ext}}\left(\frac{1}{P_2} - \frac{1}{P_1}\right) = 4.3 \times 0.29304 \approx 1.2601$

Hence,

$w = -12179.01 \times 1.2601 \approx -15345\,\text{J}$

So,

$q = -w \approx +15345\,\text{J}$

for 5 moles.

5. Heat per mole

$q_{\text{per mole}} = \frac{15345}{5} \approx 3069\,\text{J mol}^{-1}$

$q_{\text{per mole}} \approx 3.07\,\text{kJ mol}^{-1}$

Rounded to nearest integer:

$\boxed{3}$

6. Check physical consistency

The problem says the gas is expanded from $2.1\,\text{MPa}$ to $1.3\,\text{MPa}$. For expansion, the external pressure should normally be less than the gas pressure, but $4.3\,\text{MPa}$ is greater than both given pressures. That is physically inconsistent with expansion. This strongly suggests a likely misprint in the question, possibly the external pressure should be much smaller.

If one incorrectly uses the reversible isothermal formula,

$q = nRT\ln\frac{P_1}{P_2}$

then

$q = 5\times 8.314\times 293\times \ln\left(\frac{2.1}{1.3}\right) \approx 5844\,\text{J}$

Per mole,

$\approx 1.17\,\text{kJ mol}^{-1}$

which is still not $15$.

So the stored answer does not match the given data under standard thermodynamic interpretation.

7. Conclusion

Using the given data and the stated constant external pressure process,

$\boxed{q = 3\,\text{kJ mol}^{-1}}$

(rounded to nearest integer).