how to obtain the plaintext in counter mode using the AES block cipher
You can't obtain the nonce without having the key which was used for the block cipher.

For CTR mode the ciphertext is $c = m \oplus s$ where $m$ is the message and $s$ is the keystream (which is derived from the key $k$, a nonce and a counter, but this is a minor detail, important is, that re-using the nonce results in the same keystream).

Take two messages $m_1$ and $m_2$ and their corresponding ciphertexts $c_1$ and $c_2$. If we XOR those ciphertexts we get: $c_1 \oplus c_2 = m_1 \oplus s \oplus m_2 \oplus s = m_1 \oplus m_2 \oplus s \oplus s = \mathrm{???}$. Just apply the rules of XOR and you will see, what you get.

Our exercise sheet you are trying to solve, provided another hint: XORing a " " (space) with a letter (in ASCII) flips between lower- and uppercase, e.g. $\mathrm{‘a\verb!’!} \oplus \mathrm{‘\;\verb!’!} = \mathrm{‘A\verb!’!}$. If you XOR two plaintext messages in a character position, where one plaintext message has a space, what can you tell about the other message if you see a letter in uppercase or lowercase?

Now, you might ask yourself: How do I figure out, which of the two plaintexts contains the space? Use the other ciphertexts (there are 11 of them on the task sheet with the same nonce!). What will most of them tell you, if you XOR them with a ciphertext, you presume to have a space at position $i$?

Looking forward to see your solution tomorrow!

Your friendly teaching assistant

PS: You can ask on our Mattermost chat too ;-)