Solution:

The last two base-$143$ digits are $01$ exactly when $3^n \equiv 1 \pmod{143^2},$ and since $143^2 = 11^2 \cdot 13^2,$ this holds exactly when $3^n \equiv 1$ modulo both $11^2$ and $13^2.$

Modulo $121:$ $3^5 = 243 = 2 \cdot 121 + 1 \equiv 1,$ and since $5$ is prime and $3 \not\equiv 1,$ the order of $3$ is exactly $5.$ Modulo $169:$ the order of $3$ modulo $13$ is $3,$ so the order modulo $169$ is a multiple of $3.$ Writing $3^3 = 27 = 1 + 26$ and noting $26^2 = 4 \cdot 169 \equiv 0 \pmod{169},$ the binomial theorem gives $3^{3k} = (1 + 26)^k \equiv 1 + 26k \pmod{169},$ which is $1$ exactly when $13 \mid k.$ So the order of $3$ modulo $169$ is $39.$

Therefore $n$ must be a common multiple of $5$ and $39,$ and the least is $\operatorname{lcm}(5, 39) = 195.$