1. Use Kepler’s third law for planets revolving around the Sun:

$T^2 \propto r^3$

where $T$ is the time period and $r$ is the orbital radius.

2. Initially, for Earth:

$r_1 = R, \quad T_1 = 1 \text{ year}$

If the distance becomes $3R$:

$r_2 = 3R$

3. Apply Kepler’s law in ratio form:

$\left(\frac{T_2}{T_1}\right)^2 = \left(\frac{r_2}{r_1}\right)^3$

Substitute the values:

$\left(\frac{T_2}{1}\right)^2 = \left(\frac{3R}{R}\right)^3 = 3^3 = 27$

So,

$T_2^2 = 27$

$T_2 = \sqrt{27} = 3\sqrt{3} \text{ years}$

4. Therefore, the correct option is:

$\boxed{\text{D: } 3\sqrt{3} \text{ years}}$

5. Comparison with stored correct answer:

Stored correct answer is D, which matches our derived answer.