1. Use the capacitor discharge relation

When a charged capacitor discharges through a resistance $R$, the charge, voltage, and electric field all decay exponentially:

$E = E_0 e^{-t/RC}$

Given that the electric field becomes one-third of its initial value in time $t = 6.6\,\mu s$,

$\frac{E}{E_0} = \frac{1}{3} = e^{-t/RC}$

So,

$e^{-t/RC} = \frac{1}{3}$

Taking logarithm,

$\frac{t}{RC} = \ln 3$

Hence,

$R = \frac{t}{C\ln 3}$

2. Substitute the given values

Capacitance:

$C = 1.5\,\mu F = 1.5 \times 10^{-6} F$

Time:

$t = 6.6\,\mu s = 6.6 \times 10^{-6} s$

Given $\log 3 = 1.1$. In such JEE problems, this means common logarithm, so:

$\ln 3 = 2.303 \log 3 = 2.303 \times 1.1 \approx 2.53$

Now,

$R = \frac{6.6 \times 10^{-6}}{(1.5 \times 10^{-6})(2.53)}$

$R = \frac{6.6}{1.5 \times 2.53}$

$R = \frac{6.6}{3.795} \approx 1.74\,\Omega$

This does not match the stored answer, so let us check the intended interpretation.

3. Using the exam convention with base-10 logarithm directly

From

$\frac{1}{3} = e^{-t/RC}$

Taking common log on both sides,

$\log\left(\frac{1}{3}\right) = -\frac{t}{RC}\log e$

$-\log 3 = -\frac{t}{RC}(0.434)$

So,

$R = \frac{0.434\, t}{C\log 3}$

Substitute values:

$R = \frac{0.434 \times 6.6 \times 10^{-6}}{1.5 \times 10^{-6} \times 1.1}$

$R = \frac{0.434 \times 6.6}{1.65}$

$R \approx 1.736\,\Omega$

Again, this gives about $1.74\,\Omega$.

4. Possible intended textbook approximation

If one incorrectly uses

$\ln 3 \approx 1.1$

then

$R = \frac{6.6\times 10^{-6}}{1.5\times 10^{-6}\times 1.1} = \frac{6.6}{1.65} = 4\,\Omega$

This matches the stored answer.

5. Conclusion

Mathematically correct evaluation gives:

$R \approx 1.74\,\Omega$

But the stored answer $4$ is obtained only if one uses $\ln 3 = 1.1$, whereas the problem explicitly gives $\log 3 = 1.1$, which is itself inconsistent with standard logarithm values.

So the question/data appears flawed, and the intended answer is likely $4\,\Omega$ based on the setter's approximation.