1. **Stating the problem:** We want to find the value of $a$ such that the expression $M = 2^3 \times 3^2 \times a$ is a perfect cube. 2. **Recall the rule for perfect cubes:** A number is a perfect cube if all prime factors have exponents that are multiples of 3. 3. **Analyze the given expression:** $$M = 2^3 \times 3^2 \times a$$ 4. **Prime factorization of $a$:** Let $a = 2^x \times 3^y \times \text{(other primes)}$. 5. **Combine exponents:** $$M = 2^{3+x} \times 3^{2+y} \times \text{(other primes)}$$ 6. **Conditions for perfect cube:** - $3 + x$ must be divisible by 3. - $2 + y$ must be divisible by 3. - Other primes in $a$ must have exponents divisible by 3 (assume none for simplicity). 7. **Check the options:** - A) $a=3$: prime factorization $3^1$. - $x=0$, $y=1$. - $3+x=3$ divisible by 3. - $2+y=3$ divisible by 3. - So $M$ is a perfect cube. - B) $a=4=2^2$: - $x=2$, $y=0$. - $3+2=5$ not divisible by 3. - C) $a=7$ (prime): - introduces a prime factor 7 with exponent 1, not divisible by 3. - D) $a=8=2^3$: - $x=3$, $y=0$. - $3+3=6$ divisible by 3. - $2+0=2$ not divisible by 3. 8. **Conclusion:** Only option A ($a=3$) makes $M$ a perfect cube. **Final answer:** $\boxed{3}$