Any multiple of $10^{44}$ is of the form $10^{44} a$, where $a \in \mathbb{Z}$. We want $10^{44} a \vert 10^{55}$. Equivalently, we want $a \vert 10^{11}$. So all you need is to count the number of divisors of $10^{11}$. Can you complete it from here?
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In general, if you know that prime factorization of $m$, i.e. say $m= p_1^{\alpha}p_2^{\alpha_2} \cdots p_k^{\alpha_k}$, then by a simple counting argument the number of divisors of $m$ is given by (why?) $d(m) = (1+\alpha_1)(1+\alpha_2) \cdots (1+\alpha_k)$
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Hence, all you need to get the prime decomposition of $10^{11}$. The prime decomposition of $10^{11}$ is $2^{11} \times 5^{11}$. Hence, the number of divisors of $10^{11}$ is given by $(1+11) \times (1+11) = 144$. Hence, the number of multiples of $10^{44}$ that divide $10^{55}$ is $144$.