Correct Option (B)

Let Meera's age in 2002 be $x$ years.

According to the problem statement, Meenu's age in 2002 was one-third of Meera's age.

Meenu's age in 2002 = $\frac{1}{3} \times x = \frac{x}{3}$ years.

Eight years later, in 2010, their ages would be:

Meera's age in 2010 = $(x + 8)$ years.
Meenu's age in 2010 = $(\frac{x}{3} + 8)$ years.

In 2010, Meenu's age was half the age of Meera. This can be expressed as:

$\frac{x}{3} + 8 = \frac{1}{2} (x + 8)$

To solve for $x$ :

$\frac{x}{3} + 8 = \frac{x}{2} + 4$

Rearranging the terms to isolate $x$ :

$8 - 4 = \frac{x}{2} - \frac{x}{3}$

$4 = \frac{3 x - 2 x}{6}$

$4 = \frac{x}{6}$

$x = 24$

Therefore, Meera's age in 2002 was 24 years.

Meenu's age in 2002 was $\frac{1}{3} \times 24 = 8$ years.

To find Meenu's year of birth, subtract her age in 2002 from the year 2002:

Meenu's year of birth = 2002 - 8 = 1994.

Incorrect Options:

The calculated year of birth for Meenu is 1994. Options A (1992), C (1996), and D (1998) are inconsistent with the derived ages based on the given conditions.