1. Total number of 5-digit numbers

A 5-digit number ranges from $10000$ to $99999$. So the total number of 5-digit numbers is $99999-10000+1=90000.$

2. We want numbers made from exactly two digits

This means in the 5-digit number, only two distinct digits appear.

We count such numbers carefully.

3. Case 1: One of the digits is $0$

Let the two digits be $\{0,a\}$ where $a\in\{1,2,\dots,9\}$. So there are $9$ choices for $a$.

Now form 5-digit numbers using only $0$ and $a$, with both digits appearing, and the first digit cannot be $0$.

Since the first digit cannot be $0$, it must be $a$. The remaining 4 positions can each be $0$ or $a$: $2^4=16$ ways.

But the number $aaaaa$ uses only one distinct digit, so exclude it. Thus for each $a$, valid numbers are $16-1=15.$

Hence total from this case: $9\times 15=135.$

4. Case 2: Both digits are nonzero

Let the two digits be chosen from $\{1,2,\dots,9\}$. Number of ways to choose 2 digits: $\binom{9}{2}=36.$

For a fixed pair of digits, each of the 5 positions can be filled in 2 ways: $2^5=32.$

But exclude the two numbers where all digits are the same, since we need exactly two distinct digits: $32-2=30.$

So total from this case: $36\times 30=1080.$

5. Total favorable numbers

$135+1080=1215.$

6. Required probability

$P=\frac{1215}{90000}.$

Simplify: $\frac{1215}{90000}=\frac{27}{2000}=\frac{135}{10000}.$

So the probability is $\boxed{\frac{135}{10^4}}.$

7. Compare with options

Option D is $\frac{135}{10^4},$ which matches our result.