It doesn't look possible because you still need the private key to compute s since s = k−1(e + rdU) mod n. where both k and du here would be the private key which you don't have.
Hello
no possible
but cannot reach the desired result.
Let me show you the possible result.
r = 1ca4aa8c1bec706e817e9d74b356bcab13625061c541052ddd9e6352cba6911e
s = 95acfc5533fc1693918a0ac0f03c62a2b0b1d30289d8769c66b37c2e85abbcbd
z = 4bc17560a03e004105d48f22ecb8d7b620f0ba85d6619abcffad00a839216e9a
pubkey = 023d9bc5aec4e53f59b03bc4866453a94b673e99f67bd69d2915a39964d4918a98
thank you.