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(I made my answer community wiki because the previous comments cover the important points of the answer.)

You can find the relationship between inaccessible cardinals and Grothendieck universes on Wikipedia (!)

Theorem. The followings are equivalent:

1. Tarski's axiom A: every set is contained in a Grothendieck universe, and

2. There is a proper class of inaccessible cardinals.

Especially, Tarski-Grothendieck set theory $\mathsf{TG}$ and $\mathsf{ZFC}+$"There is a proper class of inaccessibles" are the same theory. It follows from the fact that every Grothendieck universe is of the form $V_\kappa$ for some inaccessible $\kappa$, where $V_\alpha$ is the $\alpha$th cumulative hierarchy. (See Trever Wilson's previous answer for the detailed proof.)

Since being inaccessible is downward absolute between $V$ and $L$, we have $$L\models \text{there is a proper class of inaccessible cardinals}$$ if $V$ has a proper class of inaccessibles. (Thank you for Noah Schweber to point it out. Being inaccessible need not be upward absolute.) It means $\mathsf{TG}+(V=L)$ is consistent if $\mathsf{TG}$ were.

(I made my answer community wiki because the previous comments cover the important points of the answer.)

You can find the relationship between inaccessible cardinals and Grothendieck universes on Wikipedia (!)

Theorem. The followings are equivalent:

1. Tarski's axiom A: every set is contained in a Grothendieck universe, and

2. There is a proper class of inaccessible cardinals.

Especially, Tarski-Grothendieck set theory $\mathsf{TG}$ and $\mathsf{ZFC}+$"There is a proper class of inaccessibles" are the same theory. It follows from the fact that every Grothendieck universe is of the form $V_\kappa$ for some inaccessible $\kappa$, where $V_\alpha$ is the $\alpha$th cumulative hierarchy. (See Trevor Wilson's previous answer for the detailed proof.)

Since being inaccessible is downward absolute between $V$ and $L$, we have $$L\models \text{there is a proper class of inaccessible cardinals}$$ if $V$ has a proper class of inaccessibles. (Thank you for Noah Schweber to point it out. Being inaccessible need not be upward absolute.) It means $\mathsf{TG}+(V=L)$ is consistent if $\mathsf{TG}$ were.

(I made my answer community wiki because the previous comments cover the important points of the answer.)

You can find the relationship between inaccessible cardinals and Grothendieck universes on Wikipedia (!)

Theorem. The followings are equivalent:

1. Tarski's axiom A: every set is contained in a Grothendieck universe, and

2. There is a proper class of inaccessible cardinals.

Especially, Tarski-Grothendieck set theory $\mathsf{TG}$ and $\mathsf{ZFC}+$"There is a proper class of inaccessibles" are the same theory. It follows from the fact that every Grothendieck universe is of the form $V_\kappa$ for some inaccessible $\kappa$, where $V_\alpha$ is the $\alpha$th cumulative hierarchy. (See Trever Wilson's previous answer for the detailed proof.)

Since being inaccessible is absolute between $V$ and $L$, we have $$L\models \text{there is a proper class of inaccessible cardinals}$$ if $V$ has a proper class of inaccessibles. It means $\mathsf{TG}+(V=L)$ is consistent if $\mathsf{TG}$ were.

(I made my answer community wiki because the previous comments cover the important points of the answer.)

You can find the relationship between inaccessible cardinals and Grothendieck universes on Wikipedia (!)

Theorem. The followings are equivalent:

1. Tarski's axiom A: every set is contained in a Grothendieck universe, and

2. There is a proper class of inaccessible cardinals.

Especially, Tarski-Grothendieck set theory $\mathsf{TG}$ and $\mathsf{ZFC}+$"There is a proper class of inaccessibles" are the same theory. It follows from the fact that every Grothendieck universe is of the form $V_\kappa$ for some inaccessible $\kappa$, where $V_\alpha$ is the $\alpha$th cumulative hierarchy. (See Trever Wilson's previous answer for the detailed proof.)

Since being inaccessible is downward absolute between $V$ and $L$, we have $$L\models \text{there is a proper class of inaccessible cardinals}$$ if $V$ has a proper class of inaccessibles. (Thank you for Noah Schweber to point it out. Being inaccessible need not be upward absolute.) It means $\mathsf{TG}+(V=L)$ is consistent if $\mathsf{TG}$ were.

(I made my answer community wiki because the previous comments cover the important points of the answer.)

You can find the relationship between inaccessible cardinals and Grothendieck universes on Wikipedia (!)

Theorem. The followings are equivalent:

1. Tarski's axiom A: every set is contained in a Grothendieck universe, and

2. There is a proper class of inaccessible cardinals.

(See Trever Wilson's previous answer for the proof.)

Especially, Tarski-Grothendieck set theory $\mathsf{TG}$ and $\mathsf{ZFC}+$"There is a proper class of inaccessibles" are the same theory.

It follows from the fact that every Grothendieck universe is of the form $V_\kappa$ for some inaccessible $\kappa$, where $V_\alpha$ is the $\alpha$th cumulative hierarchy. Since being inaccessible is absolute between $V$ and $L$, we have $$L\models \text{there is a proper class of inaccessible cardinals}$$ if $V$ has a proper class of inaccessibles. It means $\mathsf{TG}+(V=L)$ is consistent if $\mathsf{TG}$ were.

(I made my answer community wiki because the previous comments cover the important points of the answer.)

You can find the relationship between inaccessible cardinals and Grothendieck universes on Wikipedia (!)

Theorem. The followings are equivalent:

1. Tarski's axiom A: every set is contained in a Grothendieck universe, and

2. There is a proper class of inaccessible cardinals.

Especially, Tarski-Grothendieck set theory $\mathsf{TG}$ and $\mathsf{ZFC}+$"There is a proper class of inaccessibles" are the same theory. It follows from the fact that every Grothendieck universe is of the form $V_\kappa$ for some inaccessible $\kappa$, where $V_\alpha$ is the $\alpha$th cumulative hierarchy. (See Trever Wilson's previous answer for the detailed proof.)

Since being inaccessible is absolute between $V$ and $L$, we have $$L\models \text{there is a proper class of inaccessible cardinals}$$ if $V$ has a proper class of inaccessibles. It means $\mathsf{TG}+(V=L)$ is consistent if $\mathsf{TG}$ were.