1. Work done in one lift

The person lifts a mass $m=10\,\text{kg}$ through height $h=1\,\text{m}$.

So, work done in one lift is

$W_1 = mgh = 10 \times 9.8 \times 1 = 98\,\text{J}$

2. Total mechanical work done in 1000 lifts

Since the mass is lifted $1000$ times,

$W = 1000 \times 98 = 9.8 \times 10^4\,\text{J}$

We are told to consider the work done only when the weight is lifted up. So this is the useful mechanical work.

3. Relate fat energy to mechanical energy

Fat provides energy at the rate

$3.8 \times 10^7\,\text{J/kg}$

But only $20\%$ of this is converted into mechanical energy.

So, effective mechanical energy obtained from $1\,\text{kg}$ of fat is

$0.20 \times 3.8 \times 10^7 = 7.6 \times 10^6\,\text{J/kg}$

4. Fat consumed

If $M$ is the mass of fat used, then

$M = \frac{\text{required mechanical energy}}{\text{mechanical energy per kg of fat}}$

$M = \frac{9.8 \times 10^4}{7.6 \times 10^6}$

$M = \frac{9.8}{7.6} \times 10^{-2}$

$M \approx 1.289 \times 10^{-2}\,\text{kg}$

$M = 12.89 \times 10^{-3}\,\text{kg}$

5. Match with options

This corresponds to:

Option B: $12.89 \times 10^{-3}\,\text{kg}$