3. Let $p_{1}, p_{2}, \ldots, p_{m}$ be the prime divisors of $a$. Among them, by condition, there are 3 and 5, for example, $\quad p_{1}=3, p_{2}=5$. Then $a=3^{s_{1}} \cdot 5^{s_{2}} \cdot p_{3}^{s_{3}} \cdot \ldots \cdot p_{m}^{s_{m}}$ and the total number of its divisors is $\left(s_{1}+1\right)\left(s_{2}+1\right) \cdot \ldots \cdot\left(s_{m}+1\right)=63=7 \cdot 9, s_{1} \geq 1, s_{2} \geq 1, s_{i} \geq 0, i=3,4, \ldots, m$. Then the following values of pairs $\left(s_{1} ; s_{2}\right)$ are possible:

A) $s_{1}=2, s_{2}=2$ B) $s_{1}=2, s_{2}=6$, C) $s_{1}=6, s_{2}=2$, D) $s_{1}=2, s_{2}=20$,

E) $s_{1}=20, s_{2}=2$.

Case A. $s_{1}=2, s_{2}=2$ leads to the condition

$$
\left(s_{3}+1\right)\left(s_{4}+1\right) \cdot \ldots \cdot\left(s_{m}+1\right)=7
$$

For $p_{3}<p_{4}<\ldots<p_{m}$, the smallest value of $a$ corresponds to $p_{3}=2$ and $s_{3}=6$. Therefore, $a_{1}=2^{6} \cdot 3^{2} \cdot 5^{2}$.

Case B. $s_{1}=2, s_{2}=6$ leads to the condition

$$
\left(s_{3}+1\right)\left(s_{4}+1\right) \cdot \ldots \cdot\left(s_{m}+1\right)=3
$$

Therefore, $s_{3}=2$ and, consequently, $a_{2}=2^{2} \cdot 3^{2} \cdot 5^{6}$

Case C. $s_{1}=6, s_{2}=2$ leads to the condition

$$
\left(s_{3}+1\right)\left(s_{4}+1\right) \cdot \ldots \cdot\left(s_{m}+1\right)=3
$$

Therefore, $s_{3}=2$ and, consequently, $a_{3}=2^{2} \cdot 3^{6} \cdot 5^{2}$

Case D. $s_{1}=2, s_{2}=20$ leads to the condition

$$
\left(s_{3}+1\right)\left(s_{4}+1\right) \cdot \ldots \cdot\left(s_{m}+1\right)=1
$$

Therefore, $s_{3}=s_{4}=,,,=s_{m}=0$. Hence, $a_{4}=3^{2} \cdot 5^{20}$

Case E. $s_{1}=20, s_{2}=2$ leads to the condition

$$
\left(s_{3}+1\right)\left(s_{4}+1\right) \cdot,, \cdot\left(s_{m}+1\right)=1
$$

Therefore, $s_{3}=s_{4}=\cdots=s_{m}=0$. Hence, $a_{5}=3^{20} \cdot 5^{2}$.

Comparing all these cases, we arrive at the answer

$$
a_{\min }=a_{1}=2^{6} \cdot 3^{2} \cdot 5^{2}=14400
$$
Answer: $a_{\text {min }}=a_{1}=2^{6} \cdot 3^{2} \cdot 5^{2}$.