We can not apply Euler's Totient Theorem directly as $2,840$ are not relatively prime.
Let $2^{1930}=840r+s\implies s=8(2^{1927}-105r)\implies 8\mid s,s=8t$(say)
So, $2^{1927}\equiv t\pmod {105}$
Now we can safely apply Totient Theorem.
As $\phi(105)=\phi(3)\phi(5)\phi(7)=2\cdot4\cdot6=48\implies 2^{48}\equiv1\pmod{105}--->(1)$
In fact,we can potentially find some lower exponent $(e)$ of $2$ such that $2^e\equiv1\pmod{105}$ by applying Carmichael Function.
As $\lambda(105)=lcm(\lambda(3),\lambda(5),\lambda(7))$ $=lcm(\phi(3),\phi(5),\phi(7))=lcm(2,4,6)=12 \implies 2^{12}\equiv1\pmod {105}--->(2)$
So, $2^{1920}=(2^{48})^{40}\equiv1^{40}\pmod{105}=1$ by $(1)$
or $2^{1920}=(2^{12})^{160}\equiv1^{160}\pmod{105}=1$ by $(2)$
So in any case, $2^{1920}=1+105u$ for some integer $u$
$2^{1930}=2^{10}\cdot2^{1920}=2^{10}(1+105u)$ $\equiv2^{10}\pmod{2^{10}\cdot105}\equiv1024\pmod{2^{10}\cdot105}\equiv 1024\pmod{840}\equiv184$
Alternatively, $1927\equiv7\pmod {48}\equiv7\pmod {12}$
So, $2^{1927}\equiv 2^7\pmod {105}\equiv128\equiv 23$
So, $2^{1930}\equiv 2^3\cdot 23\pmod {840}\equiv 184$