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Note that: $10!$ first nonzero digit from right is 8, so if you times $1,2,\ldots,10$ the first nonzero digit will be 8, then $20!\equiv8^2\pmod{10}$ , $30!\equiv8^3\pmod{10}$ , $\ldots$ , $340!\equiv8^{34}\pmod{10}\equiv4\pmod{10}$

So the first nonzero digit from right of $340!$ is 4, then for $347!$ , it will be the first nonzero digit from right of $347*346*345*344*333*332*331*4\equiv7*6*5*4*3*2*1*4$ which is $\boxed6$