Only (A) is regular.
For (A), you can write, set $A$ as $\{1y, 11y,111y,…. \}$ where $y \in \{0+1\}^*$ and $y$ should contain at least $k$ $1s$ for every $k.$
So, $1y$ means $11,110,101,111,1001,1010,1100,1110,1101,1111,...$
$11y$ is concatenated by $1$ and elements $1y$ so elements $11y$ are $\{1111,110101,...\}$
Similarly for $111y,1111y,….$
Now, elements $1y$ starts with $1$ and having “at least” two $1s,$ It means elements $11y,111y,...$ are nothing but the some elements which are represented by $1y$
If we try to represent elements $1y$ in the form of regular expression then we should have to start with $1$ and then make zero or more than zero number of $0s$ then $1$ to make at least two $1s$ and then any number of $0s$ and $1s$. So, we can write in elements $1y$ in the form of regular expression as $10^*1(0+1)^*$
So, Regular Expression for set $A$ will be $10^*1(0+1)^*$ and hence, it is regular set.
Set $B$ is not regular set and we can prove it by pumping lemma.
Let, $B$ be a regular set and consider an element of set $B$ as $z = 1^p 001^p$ such that $|z| \geq p$ (p (pumping length) is a minimum number of states in the FSM for set B)
Now $\exists v,w,x$ such that $z = vwx= 1^{a}1^b 1^{p-a-b} 00 1^p$ where $v = 1^{a}, w=1^b, x=1^{p-a-b}001^{p}$ and $|vw| \leq p$ and $|w| \geq 1$ for all $i \geq 0$ [$p-a-b \geq 0 \implies a+b \leq p$] and $a,b \geq 0$
Now, according to pumping lemma, $\forall i \geq 0,$ $vw^ix$ should also be in $B$ .
But for $i=0,$ $z= vw^0 x = 1^{a} 1^{p-a-b}001^p = 1^{p-b}001^p$, Now here we can’t write $z$ in the form of $1^k y$ where $y \in \{0+1\}^*$ and $y$ is having at most $k$ $1s$ for every $k$ because for $b \geq 0, p-b \leq p$ Hence, $B$ is not a regular set.
Puming Lemma