1. Let the 3-digit number be $n$.

We need $\gcd(n,36)=2.$ Since $36=2^2\cdot 3^2,$ this means:

• $n$ must be divisible by $2$,
• $n$ must not be divisible by $4$ (otherwise gcd would contain $2^2$),
• $n$ must not be divisible by $3$ (otherwise gcd would also contain a factor $3$).

So we must count 3-digit numbers that are:

1. even,
2. not divisible by $4$,
3. not divisible by $3$.

2. Write $n=2m$.

Since $100\le n\le 999$, we get $50\le m\le 499.$ Also, $\gcd(n,36)=2 \iff \gcd(2m,36)=2.$ Because $36=2\cdot 18$, this requires:

• $m$ is odd (so that $n$ is not divisible by $4$),
• $m$ is not divisible by $3$.

Thus we need to count integers $m$ in $[50,499]$ such that $m$ is odd and not divisible by $3$.

3. Count odd integers from $50$ to $499$.

The odd numbers are $51,53,55,\dots,499.$ This is an arithmetic progression with common difference $2$. Number of terms: $\frac{499-51}{2}+1=224+1=225.$

So there are $225$ odd integers in this interval.

4. Among these odd integers, subtract those divisible by $3$.

An odd multiple of $3$ is a number congruent to $3 \pmod 6$. From $50$ to $499$, these are $51,57,63,\dots,495.$ This is an AP with common difference $6$. Number of terms: $\frac{495-51}{6}+1=74+1=75.$

So odd numbers not divisible by $3$ are $225-75=150.$

5. Therefore the required number of 3-digit integers $n$ is $150.$

Hence, the total number of 3-digit numbers whose greatest common divisor with $36$ is $2$ equals $150$.