This is a Bayes' theorem question. The formula for Bayes' theorem is:
$$ P(A|B) = \frac{P(B|A) \cdot P(A)}{P(B)} $$
In this case:
A = Unfair coin
B = Getting heads 10 times in a row
Hence,
$$ P(\text{Unfair coin} \mid 10 \text{ heads in a row}) = \frac{P(10 \text{ heads in a row} \mid \text{Unfair coin}) \cdot P(\text{Unfair coin})}{P(10 \text{ heads in a row})} $$
We have the following values:
$$ P(10 \text{ heads in a row} \mid \text{Unfair coin}) = 1 \quad \text{(Because both sides of the unfair coin are heads)} $$
$$ P(\text{Unfair coin}) = \frac{1}{100} $$
$$ P(10 \text{ heads in a row}) = \frac{1}{100} \cdot 1 + \frac{99}{100} \cdot \frac{1}{2^{10}} $$
Therefore,
$$ P(\text{Unfair coin} \mid 10 \text{ heads in a row}) = \frac{P(10 \text{ heads in a row} \mid \text{Unfair coin}) \cdot P(\text{Unfair coin})}{P(10 \text{ heads in a row})} $$
$$ = \frac{1 \cdot \frac{1}{100}}{\left(1 \cdot \frac{1}{100} + \frac{1}{2^{10}} \cdot \frac{99}{100}\right)} $$
$$ \approx 0.9118 \quad \text{(or approximately 91.18%)} $$
Therefore, the probability of the coin being unfair given that it landed heads 10 times in a row is approximately 91.18%.