1. Identify the required quantity

We need the dimensions of:

$\frac{\text{modulus of elasticity}}{\text{torque}}$

and it is given as:

$[M^aL^bT^c]$

with $b=3$. We must find $c$.

2. Dimension of modulus of elasticity

Modulus of elasticity has the same dimensions as stress:

$\text{Modulus} = \frac{\text{stress}}{\text{strain}}$

Since strain is dimensionless,

$[\text{Modulus}] = [\text{Stress}]$

Now,

$\text{Stress} = \frac{\text{Force}}{\text{Area}}$

Force has dimension:

$[F] = [MLT^{-2}]$

Area has dimension:

$[L^2]$

So,

$[\text{Modulus}] = \frac{[MLT^{-2}]}{[L^2]} = [ML^{-1}T^{-2}]$

3. Dimension of torque

Torque = force $\times$ distance

$[\tau] = [MLT^{-2}] \cdot [L] = [ML^2T^{-2}]$

4. Dimension of modulus per unit torque

$\left[\frac{\text{modulus}}{\text{torque}}\right] = \frac{[ML^{-1}T^{-2}]}{[ML^2T^{-2}]}$

Subtract powers:

• For $M$: $1-1=0$
• For $L$: $-1-2=-3$
• For $T$: $-2-(-2)=0$

Hence,

$\left[\frac{\text{modulus}}{\text{torque}}\right] = [M^0L^{-3}T^0]$

So,

$a=0,\quad b=-3,\quad c=0$

5. Use the given condition

The question states $b=3$, but from dimensional analysis we get $b=-3$.

This strongly suggests a sign typo in the question statement. Regardless, the time exponent is:

$c=0$

6. Final answer

$\boxed{0}$

7. Comparison with stored correct answer

Stored correct answer = $0$

Our derived answer for $c$ is also $0$, so the answer agrees.