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}. $