The $146^\text{th}$ digit of $\frac1{293}$ is
$$
d = \left\lfloor\frac{10^{146}}{293}\right\rfloor
- 10 \left\lfloor\frac{10^{145}}{293}\right\rfloor,
$$
and sage says it's 3 (sage doesn't count the leading zeros in the decimal mantissa):
(1/293).n(digits=144)
0.00341296928327645051194539249146757679180887372013651877133105802047781569965870307167235494880546075085324232081911262798634812286689419795221843
floor(10^146/293)-10*floor(10^145/293)
3
As to why, well, $p=293$ is prime, and $p-1=292=2^2\cdot73$, and any integer $a$ which is relatively prime to $p$ will have order $d$ dividing $p-1$. So the smallest positive power $d$ of $a=10$ so that $a^d\equiv1\pmod p$ must be $2,4,73,2\cdot73=146$ or, if none of these, then $4\cdot73=292$. Now $d=2$ and $d=4$ can easily be ruled out since $100,1000\not\equiv1\pmod{292}$. For $d=73$, note that $73=(10010001)_2=2^6+2^3+2^0$, so that we can calculate $10^{73}$ modulo $293$ (and its square if necessary) as follows:
$$10^2=100$$
$$10^4=(100)^2=10000\equiv38\pmod{293}$$
$$10^8\equiv(38)^2=1444\equiv272\equiv-21\pmod{293}$$
$$10^9\equiv10\cdot(-21)=-210\equiv83\pmod{293}$$
$$10^{18}\equiv(83)^2=6889\equiv150\pmod{293}$$
$$10^{36}\equiv(150)^2=22500\equiv232\equiv-61\pmod{293}$$
$$10^{72}\equiv(-61)^2=3721\equiv205\equiv-88\pmod{293}$$
$$10^{73}\equiv10\cdot(-88)=-880\equiv292\equiv-1\pmod{293}$$
Since squaring the last quantity gives $1$ modulo $293$,
we find that $d=\text{ord}_{293}{10}=2\cdot73=146$.
This method is called repeated squaring: starting with $a=10$ (step $0$),
repeatedly square the result, multiplying again by $a$ (modulo $p$)
at each intermediate step $i$ (after squaring) if bit $i$,
corresponding to $2^i$ in the binary expansion of $d$, is one.
Now with @m-k's post, we see why. If $10^{146}=q\cdot293+r$,
with $q$ and $r$ given by the division algorithm, i.e.
$q,r\in\mathbb{Z}$ with $0\leq r<293$, then $q$ is the first quantity
in the formula above for $d$:
$$
q=\left\lfloor\frac{10^{146}}{293}\right\rfloor
=\frac{10^{146}-r}{293},
$$
and its last digit -- that is, its remainder modulo $10$ --
is equal to $d$:
$$
d\equiv q\pmod{10}.
$$
However, modulo $10$, we have
$$
293q\equiv-r
\pmod{10}
\quad
\implies
\quad
q\equiv293^{-1}\cdot-r
\equiv-3^{-1}r
\equiv-7r
\equiv3r
\pmod{10}.
$$
But we already found that $r=1$, since $10^{146}\equiv1\pmod{293}$,
so that
$$
d\equiv
q\equiv
3r\equiv
3\pmod{10}.
$$