how to compute IV with AES CTR if key, plaintext and ciphertext are known? Given $E_k$, which a block-cipher using the (known) key $k$, and a plaintext-ciphertext pair $(p,c)$ where $c$ is the encryption of $p$ using CTR mode, can the initialization vector be recovered using $IV = E_k^{-1}(c \oplus p)$?
Correct.

Note that normally we use the term nonce rather than IV, because the only requirement for the nonce is that it never be repeated, while an IV (e.g., for CBC) must additionally be unpredictable in advance.  Also usually you encrypt multiple 128-blocks in a single message by using E(nonce || 0, K), E(nonce || 1, K), E(nonce || 2, K), etc., as the pad, so you might get a nonce and a block counter.